Question: Multiply the following complex numbers: $({5-2i}) \cdot ({2-2i})$
Answer: Complex numbers are multiplied like any two binomials. First use the distributive property: $ ({5-2i}) \cdot ({2-2i}) = $ $ ({5} \cdot {2}) + ({5} \cdot {-2}i) + ({-2}i \cdot {2}) + ({-2}i \cdot {-2}i) $ Then simplify the terms: $ (10) + (-10i) + (-4i) + (4 \cdot i^2) $ Imaginary unit multiples can be grouped together. $ 10 + (-10 - 4)i + 4i^2 $ After we plug in $i^2 = -1$ , the result becomes $ 10 + (-10 - 4)i - 4 $ The result is simplified: $ (10 - 4) + (-14i) = 6-14i $